1. CS 290H Lecture 3 Fill: bounds and heuristics
Class time will henceforth be 3:15 to 4:30
Homework 1 due Thursday 14 Oct by 3pm
turnin file1 file2 … Makefile-or-README
Temporary bug: “turnin” only works from csil.cs.ucsb.edu
Read GLN section 4.3 and chapter 7 (you can skip 7.2.3, 7.2.4, and 7.3.3)

2. Graphs and Sparse Matrices: Cholesky factorization
Fill: new nonzeros in factor 10 1 3 2 4 5 6 7 8 9 10 1 3 2 4 5 6 7 8 9
Symmetric Gaussian elimination:
for j = 1 to n add edges between j’s higher-numbered neighbors
G(A)
G+(A) [chordal]

3. Path lemma (GLN Theorem 4.2.2)
Let G = G(A) be the graph of a symmetric, positive definite matrix, with vertices 1, 2, …, n, and let G+ = G+(A) be the filled graph.
Then (v, w) is an edge of G+ if and only if G contains a path from v to w of the form (v, x1, x2, …, xk, w) with xi < min(v, w) for each i.
(This includes the possibility k = 0, in which case (v, w) is an edge of G and therefore of G+.)

4. The (2-dimensional) model problem
Graph is a regular square grid with n = k^2 vertices.
Corresponds to matrix for regular 2D finite difference mesh.
Gives good intuition for behavior of sparse matrix algorithms on many 2-dimensional physical problems.
There’s also a 3-dimensional model problem.

5. Permutations of the 2-D model problem
Theorem: With the natural permutation, the n-vertex model problem has (n3/2) fill.
Theorem: With any permutation, the n-vertex model problem has (n log n) fill.
Theorem: With a nested dissection permutation, the n-vertex model problem has O(n log n) fill.

6. Nested dissection ordering
A separator in a graph G is a set S of vertices whose removal leaves at least two connected components.
A nested dissection ordering for an n-vertex graph G numbers its vertices from 1 to n as follows:
Find a separator S, whose removal leaves connected components T1, T2, …, Tk
Number the vertices of S from n-|S|+1 to n.
Recursively, number the vertices of each component: T1 from 1 to |T1|, T2 from |T1|+1 to |T1|+|T2|, etc.
If a component is small enough, number it arbitrarily.
It all boils down to finding good separators!